Question

What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point

Answer 1

Answer: The amount of energy absorbed by water is 5390 Calories

Explanation:

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

$q=m\times L_{vap}$

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

$L_{vap}$ = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

$q=10g\times 539Cal/g=5390Cal$

Hence, the amount of energy absorbed by water is 5390 Calories